Prove that x2 + y2 = a2 + b2.

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If x = a cos θ + b sin θ and y = a sin θ – b cos θ, prove that x2 + y2 = a2 + b2.

Answer :

x = a cos θ + b sin θ …(1)

y = a sin θ – b cos θ …(2)

By squaring and adding both the equations

x2 + y2 = (a cos θ + b sin θ)2 + (a sin θ – b cos θ)2

(a + b)2 = a2 + b2 + 2ab and (a – b)2 = a2 + b2 – 2ab

= [(a cosθ)2 + (b sinθ)2 + 2 (a cosθ) (b sinθ)] + [(a sinθ)2 + (b cosθ)2 – 2 (a sinθ) (b cosθ)]

= a2 cos2θ + b2 sin2θ + 2 ab sinθ cosθ + a2 sin2θ + b2 cos2θ – 2 ab sinθ cosθ

= a2 cos2θ + b2 sin2θ + a2 sin2θ + b2 cos2θ

= a2 (cos2θ + sin2θ) + b2 (sin2θ + cos2θ)

Here sin2θ + cos2θ = 1

= a2 (1) + b2 (1)

= a2 + b2

Hence, x2 + y2 = a2 + b2.

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